Question

A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod. The string makes a constant angle of 61.0 ∘ with the vertical as the ball moves at a constant speed in a horizontal circle.If it takes the ball 1.25 s to complete one revolution, what is the magnitude of the radial acceleration of the ball?

Answer 1

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

 R = L sin θ

 R = 0.8 x sin 61°

 R = 0.7 m

now, calculating at the angular velocity

$\omega =\dfrac{2\pi}{T}$

$\omega =\dfrac{2\pi}{1.25}$

  ω = 5.026 rad/s

now, radial acceleration

 a = r ω²

 a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²